Evaluate the given limit: $\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$

Question

Evaluate the given limit $$\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$

My Attempt : $$=\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$ $$=\lim_{x\to 0} \dfrac {x(\tan (2x)-2\tan (x))}{1-2\cos (2x)+\cos^2 (2x)}$$

Answer 1

Using $$\tan (2x) = \frac{2\tan x}{1-\tan^2 x}$$

So $$\lim_{x\rightarrow 0}\frac{x\tan (2x)-2x\tan x}{(1-\cos 2x)^2} = \lim_{x\rightarrow 0}\frac{2x\tan x\cdot \tan^2 x}{4\sin^4 x(1-\tan^2 x)}$$

So Using $\displaystyle \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$ and $\displaystyle \lim_{x\rightarrow 0}\frac{\tan x}{x} = 1$

So we have $$\lim_{x\rightarrow 0}\frac{x\tan (2x)-2x\tan x}{(1-\cos 2x)^2} = \frac{1}{2}.$$

Answer 2

With equivalents, there are less calculations:

First rewrite the expression as $\;\dfrac{x(\tan 2x-2\tan x)}{(1-\cos 2x)^2}$.

Now,

• it's standard that $\;1-\cos u\sim_0\dfrac{u^2}2$, so $\;(1-\cos 2x)^2\sim_0(2x)^2 $.
• Taylor's formula at order $3$ for the tangent is $\;\tan u=u+\dfrac{u^3}3+o(u^3)$, so $$\tan 2x-2\tan x=2x+\frac{8x^3}3 -\Bigl(2x+\frac{2x^3}3\Bigr)+o(x^3)=2x^3+o(x^3)\sim_0 2x^3, $$ and eventually $$\dfrac{x(\tan 2x-2\tan x)}{(1-\cos 2x)^2}\sim_0\frac{2x^4}{4x^4}=\frac12.$$